Lecture 01 Introduction to Computational Methodoloy

Chapter 1: Introduction

Learning Objectives

In this chapter, you will learn:

• Types of error
• Ways of error reduction
• Types of software used to solve computational problems using numerical methods

---

1. Error Analysis

Definition of Error

An error, $ee$, in Numerical Mathematics is the difference between the actual value (exact value) and its computed value.

If $x\ast x^{\ast }$ represents the computed value of a quantity, and the actual value is xx, then the error is:

$$e=x-x\ast e=x-x^{\ast }$$

Ways of Measuring Error

Absolute Error

$$eabs=\mid x-x\ast \mid e_{\text{abs}}=|x-x^{\ast }|$$

Relative Error

$$erel=eabs\mid x\mid e_{\text{rel}}=\frac{e_{\text{abs}}}{|x|}$$

---

Types of Error

Note

By default, if question doesn’t specify. Use Rounding method.

Round-off Error

Round-off error occurs when numbers are rounded to a limited number of decimal places.

Rules for rounding off a number:

• If the digit to be dropped is 0, 1, 2, 3, or 4, leave the next remaining digit unchanged.
• If the digit to be dropped is 5, 6, 7, 8, or 9, increase the next remaining digit by one.

Chopping Error

A number $x$ is chopped to $n$ digits when all digits beyond the $n$-th digit are discarded without changing any of the remaining $n$ digits.

Truncation Error

Truncation error is the error arising when an infinite series is replaced by a finite number of terms.

Example

Example: The infinite Taylor series for $ex{e}^x$ is:

ex=1+x+x22!+x33!+x44!+…e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots

$$Atruncatedversionmightincludeonlythefirst5terms.$$

---

Info

If question doesn’t specify, use 4 decimal places.

Example 1

Given an actual value $x$ = 1.485642x = 1.485642 and its computed value $x$∗=1.492101x${}^{\ast }$ = 1.492101, find:

(a) Absolute Error

$$eabs=\mid 1.485642-1.492101\mid =0.006459e_{\text{abs}}=|1.485642-1.492101|=0.006459$$

(b) Relative Error

$$erel=eabs\mid x\mid =\mathrm{0.0064591.485642}=0.00434762e_{\text{rel}}=\frac{e_{\text{abs}}}{|x|}=\frac{0.006459}{1.485642}=0.00434762$$

---

Example 2: Round-off Error

Round off $x=227x=\frac{22}{7}$ to five decimal places and find its absolute error.

$$x\ast =3.14286$$ $$0.00000289$$

---

Example 3: Chopping Error

Chop $x=227x=\frac{22}{7}$ to five decimal places and find its relative error.

$$x\ast =3.14285$$ $$e_{rel}=\frac{|x-x^{\ast }|}{|x|}=2.2727\times 10^{-}6=0.000002$$

---

2. Error Reduction

Notes

actual value means plucking in the value without rounding

Nested Form

A polynomial function is given by:

$$p(x)=a0+a1x+a2x2+\cdots +anxnp(x)=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n$$

Rewriting in nested form:

$$p(x)=a0+x(a1+x(a2+\dots (an-1+xan)))p(x)=a_0+x(a_1+x(a_2+\dots (a_{n-1}+x{a}_n)))$$

Example 4: Consider the polynomial function:

$$f(x)=6x3-3x2+2x+1.5,wherex=4.71f(x)=6x^3-3x^2+2x+1.5,\text{where }x=4.71$$

(a) Rewrite $f(x)$ in the nested form

$$\begin{array}{r}f(x)=((6x-3)x+2)x+1.5f(x)\\ =((6x-3)x+2)x+1.5\end{array}$$

(b) Compute $f(x)f(x)$ using:

(i) Exact Evaluation

$$\begin{array}{r}polynomialfunction,f(x)=x^3-6.1x^2\\ f(4.71)=(4.71{)}^3-6.1(4.71{)}^2+3.2(4.71)+1.5=104-6.1(22.2)\end{array}$$

(ii) Three-digit rounding-off evaluation (iii) Three-digit chopping evaluation

$$e_{rel}=\mid fexact-fapprox\mid \mid fexact\mid e_{\text{rel}}=\frac{|f_{\text{exact}}-f_{\text{approx}}|}{|f_{\text{exact}}|}$$

Accuracy loss due to rounding and chopping errors can be reduced using nested form.

---

3. Avoiding Loss of Significance in Subtraction

Loss of significance occurs when nearly equal numbers are subtracted.

Example: Consider two nearly equal numbers:

$$\begin{array}{r}p=3.1415926,q=3.1415957p=3.1415926\\ q=3.1415957p-q=-0.00000308p-q=-0.00000308\end{array}$$

This result has only five significant digits.

Techniques to reduce loss of significance:

• Rationalization
• Taylor series expansion

---

Example 5: Avoiding Loss of Significance

Given the function:

f(x)=x−x+1f(x) = x - \sqrt{x+1}

(a) Approximate f(500)f(500) using direct computation:

$$f(500)=500-501=500-22.3830=11.1500f(500)=500-\sqrt{501}=500-22.3830=11.1500$$

(b) Rewrite to Avoid Subtraction

$$f(x)=x2-(x+1)x+x+1f(x)=\frac{x^2-(x+1)}{x+\sqrt{x+1}}$$

(c) Approximate f(500)f(500) using the rewritten function

$$f(500)=500500+501=11.1748f(500)=\frac{500}{500+\sqrt{501}}=11.1748$$

(d) Compare Errors

$$e_{abs}=\mid 11.1748-11.1500\mid =0.0248e_{\text{abs}}=|11.1748-11.1500|=0.0248$$

Using the rewritten function reduces the loss of significance.

---

4. The Use of Taylor Series

Taylor series expansions can help avoid subtraction errors.

$$\begin{array}{r}sinx=x-x33!+x55!-\dots \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \\ tanx=x+x33+2x515+\dots \tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\dots \end{array}$$

Example 6: Given

$f(x)=tanx-sinxf(x)=\tan x-\sin x$

Rewrite using the Taylor series:

$$f(x)=x+x33-(x-x33!)f(x)=x+\frac{x^3}{3}-\left(x-\frac{x^3}{3!}\right)f(x)=x32f(x)=\frac{x^3}{2}$$